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Jul 29 at 16:57 vote accept Zhaoting Wei
Jul 29 at 13:23 history edited David Gao CC BY-SA 4.0
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Jul 29 at 8:49 history edited David Gao CC BY-SA 4.0
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Jul 29 at 8:45 comment added David Gao @AliTaghavi I just realized, funnily enough, that my proof implies $A_q$ are all isomorphic to $\mathbb{K}(\ell^2) \oplus \mathbb{C}$, the unitization of the algebra of compact operators, for any $0 < q < 1$. So I suppose that answers your previous question on whether these algebras are isomorphic, in the positive.
Jul 29 at 8:40 history edited David Gao CC BY-SA 4.0
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Jul 29 at 6:12 history edited David Gao CC BY-SA 4.0
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Jul 28 at 17:31 history edited David Gao CC BY-SA 4.0
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Jul 28 at 17:23 history edited David Gao CC BY-SA 4.0
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Jul 28 at 17:03 comment added David Gao @AliTaghavi You are certainly free to add more relations, you just get a quotient of $A_q$, that's all. That's not what the OP asked, nor do I find any reason why adding more (completely random, it seems) relations is interesting.
Jul 28 at 17:01 comment added David Gao @AliTaghavi I'm not sure what your point is. The third condition is certainly redundant, but including it makes the relations more symmetric and aesthetically pleasing, at least to me. Whether it is optimal or not, I honestly couldn't care less.
Jul 28 at 16:56 comment added David Gao To OP: The proof is now complete, with the remaining cases resolved and a concrete description of $A_q$ provided as a side product.
Jul 28 at 16:55 history edited David Gao CC BY-SA 4.0
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Jul 28 at 16:44 history edited David Gao CC BY-SA 4.0
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Jul 28 at 14:32 comment added Ali Taghavi yes I was mistaken that was an ewxercise in rudin functional analysis. i am sorry. But by the above comment I was actually thinking to the maximality and optimality of the set of relations. That was a random relation that I pointed out to(which was wrong. Ok replace it with another one say $-log(q)ab$. BTW the optimality requirement implies that we exclude the third relation in OP question. do you agree?
Jul 28 at 14:13 comment added David Gao @AliTaghavi The relation is not satisfiable, since for any bounded $a, b$, $[a,b]$ cannot be a nonzero scalar.
Jul 28 at 14:08 comment added Ali Taghavi +1 for this interesting answer. honestly I am not familiar with "universal algebra generated by elements and relations" but I am curious since 1 decade. For example what would we lose or gain if we add extra relation $[a,b]=-log(q)$ in this or in any other universal algebra? because this condition consist with any commutative algebra in a deformation of $A_q$ with q close to 1
Jul 28 at 13:46 history answered David Gao CC BY-SA 4.0